Throw an exception in the finally block

Question

I've tried to throw the same excpetion in a finally block, while the previously throwed expcetion was not catched. I expected that we have two object of Excpetion type that shal be thrown. Since we need in two catch clauses as the following:

public static void main(String[] args) {
    try {
        try {
            try {
                throw new Exception();
            } finally {
                System.out.println("finally");
                throw new Exception();
            }
        } catch (Exception ex) {
            System.out.println("catch");
        } finally {
            System.out.println("finally");
        }
    } catch (Exception ex) {
        System.out.println("catch");
    }
    System.out.println("finish");
}

But that program prints:

finally
catch
finally
finish

That is, the second catch clause was not entered. Why?

possible duplicate of [Double exception throwing in a try / finally block](http://stackoverflow.com/questions/12641834/double-exception-throwing-in-a-try-finally-block) — Jens Schauder, Aug 23 '14 at 18:12
possible duplicate of [What happens if both catch and finally blocks throw exception?](http://stackoverflow.com/questions/1482395/what-happens-if-both-catch-and-finally-blocks-throw-exception) — Rahul Tripathi, Aug 23 '14 at 18:14

Jens Schauder · Accepted Answer · 2014-08-23T18:19:59.350

4

When you throw an exception in the finally block, the first exception silently disappears.

It's in the JLS Chapter 14.20.2

If the finally block completes abruptly for reason S, then the try statement completes abruptly for reason S.

This is true how ever you entered the finally block. If you entered it by throwing an exception T that exception can not be catched anymore.

edited Aug 23 '14 at 18:19

answered Aug 23 '14 at 18:11

Jens Schauder

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Clear answer. Thank you! – Aug 23 '14 at 18:21

score 0 · Answer 2 · answered Aug 23 '14 at 18:10

0

When you throw an exception from the finally block, it supresses any exception thrown from the try block, therefore there was only one exception to catch.

The first catch cluase already caught that exception, so the second one had nothing to catch.

answered Aug 23 '14 at 18:10

Eran

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score 0 · Answer 3 · answered Aug 23 '14 at 18:12

0

Your first try-catch is already trying to catch it, the extra try blocks aren't there for any particular reason. If you try to throw more exceptions you'll notice that you'll get a syntax error, Unreachable code

Basically, keep it in one try block

answered Aug 23 '14 at 18:12

Juxhin

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Throw an exception in the finally block

3 Answers3